Electric field at the center of a semicircle

  • Thread starter rocapp
  • Start date
  • #1
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Homework Statement



See the image attached

Homework Equations


Ex = (Kλ/R)*∫3/2ππ/2(cosθ)dθ


The Attempt at a Solution



Ey cancels because there is an equal amount of force in both y directions. For the electric field on the x axis, the above equation can be used.

Given that λ=Q/L

λ=(85x10^-9 C)/(0.1 m)
=8.5x10^-7

R = L/θ
=(0.1 m)/(3/2π - π/2)
=0.03183

E = ((8.85x10^-9)(8.5x10^-7)/0.03183)*pi
= 7.424640507894822×10^-13

This is not correct. Please help! Thanks in advance.
 

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Answers and Replies

  • #2

Homework Statement



See the image attached

Homework Equations


Ex = (Kλ/R)*∫3/2ππ/2(cosθ)dθ


The Attempt at a Solution



Ey cancels because there is an equal amount of force in both y directions. For the electric field on the x axis, the above equation can be used.

Given that λ=Q/L

λ=(85x10^-9 C)/(0.1 m)
=8.5x10^-7

R = L/θ
=(0.1 m)/(3/2π - π/2)
=0.03183

E = ((8.85x10^-9)(8.5x10^-7)/0.03183)*pi
= 7.424640507894822×10^-13

This is not correct. Please help! Thanks in advance.
take an element dx at θ with width dθ
then integrate from 0 to∏
 
  • #3
95
0
Haha! I just realized my mistake as soon as you posted! Thank you very much!

To be clear for anyone who looks at this post for help in the future,

Ex = (Kλ/R)*∫3/2ππ/2(cosθ)dθ

Since the integral from pi/2 to 3pi/2 is not going to have the correct coordinates to give you the correct length, you must integrate from 0 to pi since the shape is a semicircle (half a circle) or from 0 to pi on a trigonometric circle.

When you integrate from 0 to pi/2, the charge for half the semicircle is 240072.2588752749

Multiply that number by two, and the correct answer is:

4.9×105 N/C


Thanks again!
 

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